Problem: $g(n)=80\cdot\left(\dfrac{3}{4}\right)^{{\,n}}$ Complete the recursive formula of $g(n)$. $g(1)=$
Explanation: $g( 1)=80\cdot\left(\dfrac34\right)^{ 1}={60}$ $g( 2)=80\cdot\left(\dfrac34\right)^{ 2}={45}$ $\dfrac{g( 2)}{g( 1)}=\dfrac{{45}}{{60}}={\dfrac34}$ So the first term of the sequence is ${60}$ and the common difference is ${\dfrac34}$. This is the recursive formula of the sequence: $\begin{cases} g(1)={60} \\\\ g(n)=g(n-1)\cdot {\dfrac34} \end{cases}$